Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<cmath>
#include<string>

#define ls (t<<1)
#define rs ((t<<1)+1)
#define mid ((l+r)>>1)
#define fi first
#define se second
#define mk make_pair
#define pb push_back

#define N 200005
#define M 200005
#define seed 23333

using namespace std;
typedef long long LL;
const LL INF=1e18;
int K,i,j,m,n,p,k,B[N],C[N],L[N],R[N];
LL A[N],Min[N*4],Add[N*4];
void build_tree(int l,int r,int t)
{
		if (l==r) Min[t]=-C[l];
		else 
		{
				build_tree(l,mid,ls); build_tree(mid+1,r,rs);
				Min[t]=min(Min[ls],Min[rs]);
		}
}
void down(int t)
{
		if (Add[t]==-1) return;
		Add[ls]=Add[rs]=Min[ls]=Min[rs]=INF;
		Add[t]=-1;
}
long long FindMin(int ll,int rr,int l,int r,int t)
{
		if (ll>rr) return INF;
		else if (ll<=l&&r<=rr) return Min[t];
		else 
		{
					down(t);
					long long S=INF;
					if (ll<=mid) S=min(S,FindMin(ll,rr,l,mid,ls));
					if (rr>mid)  S=min(S,FindMin(ll,rr,mid+1,r,rs));
					return S; 
		}
}
void update(int ll,long long c,int l,int r,int t)
{
		if (l==r) Min[t]=min(Min[t],c-C[l]);
		else 
		{
				down(t);
				if (ll<=mid) update(ll,c,l,mid,ls);
				else update(ll,c,mid+1,r,rs);
				Min[t]=min(Min[ls],Min[rs]);
		}
}
long long DFS(int l,int r,int t)
{
		if (l==r) return Min[t]+C[l];
		else 
		return down(t),min(DFS(l,mid,ls),DFS(mid+1,r,rs));
}
void del(int ll,int rr,int l,int r,int t)
{
		if (ll>rr) return;
		if (ll<=l&&r<=rr) Add[t]=Min[t]=INF;
		else
		{
				down(t);
				if (ll<=mid) del(ll,rr,l,mid,ls);
				if (rr>mid)  del(ll,rr,mid+1,r,rs);
				Min[t]=min(Min[ls],Min[rs]); 
		}	
}
int main()
{
		scanf("%d%d",&n,&K);
		memset(Add,-1,sizeof(Add));
		for (i=1;i<=n;++i)
		{
				scanf("%lld%d",&A[i],&B[i]);
				if (B[i]==1&&2*A[i]>K)
				{
						puts("-1");
						return 0;
				}
				A[i]+=A[i-1];
				if (B[i]==2) continue;
				L[i]=(-2*A[i-1])%K; if (L[i]<0) L[i]+=K;
				R[i]=(-2*A[i])%K; if (R[i]<0) R[i]+=K;
				C[++C[0]]=L[i]; C[++C[0]]=R[i];  
		}
		sort(C+1,C+C[0]+1); C[0]=unique(C+1,C+C[0]+1)-(C+1);
		build_tree(1,C[0],1);
		for (i=1;i<=n;++i)
		{
				if (B[i]==2) continue; 
				int l=lower_bound(C+1,C+C[0]+1,L[i])-C,r=lower_bound(C+1,C+C[0]+1,R[i])-C;
				if (l>r)
				{
						update(l,FindMin(r+1,l-1,1,C[0],1)+C[l],1,C[0],1);
						del(r+1,l-1,1,C[0],1);
				}
				else 
				{
						update(l,FindMin(r+1,C[0],1,C[0],1)+C[l]+K,1,C[0],1);
						update(l,FindMin(1,l-1,1,C[0],1)+C[l],1,C[0],1);
						del(r+1,C[0],1,C[0],1);
						del(1,l-1,1,C[0],1);
				}
		}
		printf("%lld\n",2*A[n]+DFS(1,C[0],1)); 
}